= Solution
A <Fourier mode> with $q=k^2\ge0$ evolves under $J-q\operatorname{diag}(\kappa_1,\kappa_2)$. Its <trace> remains negative for a homogeneously stable positive branch. Its <determinant> is
$$
D(q)=\kappa_1\kappa_2q^2+(s\kappa_1-u_0\kappa_2)q+s(2u_0^2-u_0).
$$
If $u_0\kappa_2\le s\kappa_1$, this is positive for every $q\ge0$. Otherwise its minimum is attained at $q_*=(u_0\kappa_2-s\kappa_1)/(2\kappa_1\kappa_2)$. Stability requires $D(q_*)>0$, or
$$
u_0\kappa_2-s\kappa_1<2\sqrt{\kappa_1\kappa_2s(2u_0^2-u_0)}.
$$
In terms of $d=\sqrt{\kappa_2/\kappa_1}$, the boundary is the positive root of $u_0d^2-s=2d\sqrt{s(2u_0^2-u_0)}$:
$$
\boxed{d_c(u_0)=\sqrt{2s}+\frac{\sqrt{s(2u_0^2-u_0)}}{u_0}.}
$$
Within $1/2<u_0<s$, all modes decay for $d<d_c$; a finite band grows for $d>d_c$, giving a <Turing instability>. Equality is neutral at one nonzero <wavenumber>.
For $s=2$, the positive lower branch $0<u_0<1/2$ is unstable for every $d$. The upper branch is homogeneously stable for $1/2<u_0<2$: it is stable to all modes to the left of $d_c=2+\sqrt{4-2/u_0}$ and unstable to spatial modes to the right. For $u_0>2$ it is already homogeneously unstable. The zero branch is stable to every spatial mode when $r<0$, because its mode <eigenvalues> are $r-\kappa_1k^2$ and $-s-\kappa_2k^2$.
\Image[/past-exam-of-the-mathematics-course-of-the-university-of-cambridge/2006/ii/paper-3-turing.png]
{title=Positive homogeneous branches and their spatial stability for s equals two}
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